Question

Coulomb's Law: Two tiny particles having

charges of +5.00 μC and +7.00 μC are

placed along the x-axis. The +5.00-μC

particle is at x = 0.00 cm, and the other

particle is at x = 100.00 cm. Where on the x-axis

must a third charged particle be placed

so that it does not experience any net

electrostatic force due to the other two

particles?

Answer 1

Charge on the first particle = q₁ = 5 C = 5 x 10^-6 C

Charge on the second particle = q₂ = 7 C = 7 x 10^-6 C

Charge on the third particle = q₃

Let the third particle be placed on the x-axis at x = d.

Distance of the third particle from the first particle = d

Distance of the third particle from the second particle = 100 - d

Electrostatic force on the third particle due to the first particle = F₁ = kq₁q₃/d²

Electrostatic force on the third particle due to the second particle = F₂ = kq₂q₃/(100-d)²

The third particle experiences zero net force,

F₁ = F₂

kq₁q₃/d² = kq₂q₃/(100-d)²

q₁/d² = q₂/(100-d)²

(5x10^-6)/d² = (7x10^-6)/(100-d)²

5/d² = 7/(100-d)²

5(100-d)² = 7d²

50000 - 1000d + 5d² = 7d²

2d² + 1000d - 50000 = 0

d² + 500d - 25000 = 0

d = 45.805 or -545.805

d is not negative.

d = 45.805 cm

The third particle should be placed at x = 45.805 cm on the x-axis so that it does not experience any net force due to the other two particles.